# What is E_"cell" for the following battery? Mg"(s)"|Mg^{2+}(0.20"M")||H_2(0.5"atm")" "H^{+}(0.300"M")|Pt"(s)"

## $M g \text{(s)"|Mg^{2+}(0.20"M")||H_2(0.5"atm")" "H^{+}(0.300"M")|Pt"(s)}$ Reaction would be: $M g \text{(s)"+2H^{+}\toMg^{2+}"(aq)"+H_2"(g)}$

Jul 26, 2018

$2.36 \text{ V}$

#### Explanation:

E_"cell"=E°_"cell"-(0.0592" V")/n\logQ, and $n = 2$ (2 ${e}^{-}$)

Based on voltaic charge table,
E°_"cell"=2.356+0.00=2.356"V" (half-reactions of magnesium and hydrogen, added together)

$\setminus \log \left(Q\right) = \setminus \log \left(\frac{\left[M {g}^{2 +}\right] {P}_{{H}_{2}}}{{H}^{+}} ^ 2\right) = \setminus \log \left(\frac{\left(0.20\right) \left(0.5 \text{atm}\right)}{0.300} ^ 2\right) \setminus \approx 0.045758$

$\setminus \therefore {E}_{\text{cell"=2.356-0.0592/2(0.045758)\approx2.356-0.00135"V"\approx2.36" V}}$

Jul 30, 2018

Well, I get ${E}_{c e l l} \approx {E}_{c e l l}^{\circ}$, because $Q \approx 1$.

I'm assuming at $\text{298.15 K}$ and $\text{1 atm}$.

In America, we write cell notation to have electrons flow from left to right. Thus, the left-hand side is the anode and right-hand side is the cathode, and we wish for $\text{Mg} \left(s\right)$ to get oxidized by ${\text{H}}^{+}$.

So, the reaction is:

$2 {\text{H"^(+)(aq) + 2e^(-) -> "H}}_{2} \left(g\right)$, ${E}_{red}^{\circ} = \text{0.00 V}$
$\underline{{\text{Mg"(s) -> "Mg}}^{2 +} \left(a q\right) + 2 {e}^{-}}$, ${E}_{red}^{\circ} = - \text{2.37 V}$
$2 {\text{H"^(+)(aq) + "Mg"(s) -> "Mg"^(2+)(aq) + "H}}_{2} \left(g\right)$

My preferred way to get ${E}_{c e l l}^{\circ}$ is:

${E}_{c e l l}^{\circ} = {E}_{red}^{\circ} + {E}_{o x}^{\circ}$

= "0.00 V" + [-(-"2.37 V")]

$= + \text{2.37 V}$

Alternatively, a no-brainer way would be to subtract the LESS positive (MORE negative) ${E}_{red}^{\circ}$ from the MORE positive (LESS negative) ${E}_{red}^{\circ}$ to guarantee the spontaneous reaction.

E_(cell)^@ = E_("cathode")^@ - E_"anode"^@

= "0.00 V" - (-"2.37 V")

$= + \text{2.37 V}$

Goodie, that was the easy part. Now since we want nonstandard, nonequilibrium conditions at $\text{298.15 K}$, we use the Nernst equation.

${E}_{c e l l} = {E}_{c e l l}^{\circ} - \frac{R T}{n F} \ln Q$

where $R$ is the universal gas constant, $T$ is temperature in $\text{K}$, $n$ is the mols of electrons PER mol of atom, $F = {\text{96485 C/mol e}}^{-}$ is Faraday's constant, and $Q$ is the reaction quotient.

Note that $\ln x \approx 2.303 \log x$.

First we get $Q$.

Q = ((["Mg"^(2+)]//c^@)(P_(H_2)//P^@))/(["H"^(+)]//c^@)^2

(where we use ${c}^{\circ} = \text{1 M}$ and ${P}^{\circ} = \text{1 atm}$ to make $Q$ unitless for use in $\ln Q$.)

= (("0.20 M"//"1 M")("0.5 atm"//"1 atm"))/("0.300 M"//"1 M")^2

$= 1.11$

Now we proceed to calculate ${E}_{c e l l}$.

color(blue)(E_(cell)) = "2.37 V" - ("8.314 V"cdotcancel"C""/"cancel"mol"cdotcancel"K" cdot 298.15 cancel"K")/((2 cancel("mol e"^(-)))/(cancel"1 mol Mg") cdot 96485 cancel"C""/"cancel("mol e"^(-))) ln (1.11)

= "2.37 V" - "0.0257 V"/("2 e"^(-)) ln (1.11)

$= \text{2.37 V" - cancel("0.0592 V"/("2 e"^(-)) log(1.11))^"small}$

$=$ $\textcolor{b l u e}{\text{2.37 V}}$ again...